Let \( y=3 x^{2} \). Find the change in \( y, \Delta y \) when \( x=4 \) and \( \Delta x=0.1 \) Find the differential \( d y \) when \( x=4 \) and \( d x=0.1 \)

To find the change in \( y \), we first substitute \( x = 4 \) into the original equation: \[ y = 3(4^2) = 3(16) = 48. \] Now, the change in \( y \) can be calculated using the derivative of \( y \) with respect to \( x \): \[ \frac{dy}{dx} = 6x. \] At \( x = 4 \), this gives: \[ \frac{dy}{dx} = 6(4) = 24. \] Now, apply the change in \( x \): \[ \Delta y = \frac{dy}{dx} \Delta x = 24(0.1) = 2.4. \] Next, we find the differential \( dy \) when \( x = 4 \) and \( dx = 0.1 \): Since the derivative \( \frac{dy}{dx} = 6x \), we again calculate at \( x = 4 \): \[ dy = \frac{dy}{dx} dx = 24(0.1) = 2.4. \] Thus, both the change in \( y \) and the differential yield the same result: \[ \Delta y \approx 2.4, \quad dy = 2.4. \]